Working with Vectors

Finding Vector Paths

Finding paths in vector diagrams

It is important to be able to describe vectors by following paths through a geometric diagram
The following grid is made up entirely of parallelograms, with the vectors and defined as marked in the diagram:

Vector parallelogram grid, IGCSE & GCSE Maths revision notes

Note the difference between "specific" and "general" vectors
- The vector $\vec{A B}$ in the diagram is specific and refers only to the vector starting at $A$ and ending at $B$
- However, the vector $a$ is a general vector - any vector the same length as $\vec{A B}$ and pointing in the same direction is equal to $a$
- Similarly, any vector the same length as $\vec{A F}$ and pointing in the same direction is equal to $b$
Following a vector in the "wrong" direction (i.e. from head to tail instead of from tail to head) makes a general vector negative
- So in the diagram above $\vec{A B} = a$ , but $\vec{B A} = - a$
- Similarly $\vec{A F} = b$ and $\vec{F A} = - b$
Note in particular the vector $\vec{F B}$ :

Getting from point to point we have to go the 'wrong way' down and then the 'right way' along

Vector paths vector FB 02, IGCSE & GCSE Maths revision notes

It follows that:
- $\vec{F B} = - b + a = a - b$
and of course then
- $\vec{B F} = - \vec{F B} = - (a - b) = b - a$

Keeping those things in mind, it is possible to describe any vector that goes from one point to another in the above diagram in terms of a and b

Exam Tip

Adding and subtracting vectors follows all the same rules as adding and subtracting letters like x and y in algebra (this includes collecting like terms).
It doesn't matter exactly what path you follow through a diagram from starting point to ending point – as long as you add and subtract the general vectors correctly along the path you use, you will get the correct answer.

Worked example

The following diagram consists of a grid of identical parallelograms.

Vectors a and b are defined by $a = \vec{A B}$ and $b = \vec{A F}$ .

Write the following vectors in terms of a and b .

\vec{A E}

To get from A to E we need to follow vector a four times to the right.

\begin{array}{rcl} \vec{A E} & = & \vec{A B} + \vec{B C} + \vec{C D} + \vec{D E} \\ = & a + a + a + a \end{array}

\vec{A E} = 4 a

\vec{G T}

There are many ways to get from G to T. One option is to go from G to Q ( b twice), and then from Q to T ( a three times).

\begin{array}{rcl} \vec{G T} & = & \vec{G L} + \vec{L Q} + \vec{Q R} + \vec{R S} + \vec{S T} \\ = & b + b + a + a + a \end{array}

\vec{G T} = 3 a + 2 b

\vec{E K}

There are many ways to get from E to K. One option is to go from E to O ( b twice), and then from O to K ( - a four times).

\begin{array}{rcl} \vec{E K} & = & \vec{E J} + \vec{J O} + \vec{O N} + \vec{N M} + \vec{M L} + \vec{L K} \\ = & b + b - a - a - a - a \end{array}

\vec{E K} = 2 b - 4 a

-4 a + 2 b also acceptable

Vector Proof

What are vector proofs?

In vector proofs we use vectors, along with a few key ideas, to prove that things are true in geometrical diagrams
Problem solving with vectors involves using these vector proofs to help us to find out additional information

Parallel vectors

Two vectors are parallel if and only if one is a multiple of the other
This tends to appear in vector proofs in the following ways:
- If you find in your workings that one vector is a multiple of the other, then you know that the two vectors are parallel – you can then use that fact in the rest of the proof
- If you need to show that two vectors are parallel, then all you need to do is show that one of the vectors multiplied by some number is equal to the other one
- E.g. If $\vec{A B} = a + 2 b$ and $\vec{C D} = 3 a + 6 b$ , then $\vec{C D} = 3 (a + 2 b) = 3 \vec{A B}$ , therefore $\vec{A B}$ and $\vec{C D}$ are parallel

Points on a straight line

Often you are asked to show in a vector proof that three points lie on a straight line (ie that they are collinear)
To show that three points $A$ , $B$ and $C$ lie on a straight line,
- show that the vectors connecting the three points are parallel,
  - for example, show that $\vec{A B}$ is a multiple of (and therefore parallel to) $\vec{A C}$ , or that $\vec{A B}$ is a multiple of (and therefore parallel) to $\vec{B C}$
- as those two vectors are parallel and they share a common point it means that the three lines form a straight line

Collinear and Non-collinear Points, IGCSE & GCSE Maths revision notes

Vectors divided in ratios

Be careful turning ratios into fractions in vector proofs!
If a point divides a line segment in the ratio p : q, then:

$\vec{A X} = \frac{p}{p + q} \vec{A B}$ and $\vec{X B} = \frac{q}{p + q} \vec{A B}$

- eg. In the following diagram, the point divides in the ratio 3: 5:

Vector line divided in ratio, IGCSE & GCSE Maths revision notes

Therefore

$\vec{A X} = \frac{3}{8} \vec{A B}$ and $\vec{X B} = \frac{5}{8} \vec{A B}$

Worked example

The diagram shows trapezium OABC.

$\vec{O A} = 2 a$

$\vec{O B} = c$

AB is parallel to OC, with $\vec{A B} = 3 \vec{O C}$ .

Question Vector Trapezium, IGCSE & GCSE Maths revision notes

Find expressions for vectors

\vec{O B}

and

\vec{A C}

in terms of a and c .

\vec{A B} = 3 \vec{O C}

and

\vec{O C} = c

\vec{A B} = 3 c

$\begin{array}{rcl} \vec{O B} & = & \vec{O A} + \vec{A B} \\ = & 2 a + 3 c \end{array}$

\vec{O B} = 2 a + 3 c

$\begin{array}{rcl} \vec{A C} & = & \vec{A O} + \vec{O C} \\ = & - \vec{O A} + \vec{O C} \\ = & - 2 a + c \end{array}$

$\vec{O B} = c - 2 a$

Point P lies on AC such that AP : PC = 3 : 1.

Find expressions for vectors

\vec{A P}

and

\vec{O P}

in terms of a and c .

AP : PC = 3 : 1 means that

\vec{A P} = \frac{3}{3 + 1} \vec{A C} = \frac{3}{4} \vec{A C}

$\begin{array}{rcl} \vec{A P} & = & \frac{3}{4} \vec{A C} = \frac{3}{4} (- 2 a + c) \end{array}$

\vec{A P} = - \frac{3}{2} a + \frac{3}{4} c

$\begin{array}{rcl} \vec{O P} & = & \vec{O A} + \vec{A P} \\ = & 2 a + (- \frac{3}{2} a + \frac{3}{4} c) \end{array}$

$\vec{O P} = \frac{1}{2} a + \frac{3}{4} c$

Hence, prove that point P lies on line OB, and determine the ratio

\vec{O P} : \vec{P B}

To show that O, P, and B are colinear (lie on the same line), note that

\vec{O P} = \frac{1}{2} a + \frac{3}{4} c = \frac{1}{4} (2 a + 3 c)

$\begin{array}{rcl} \vec{O P} & = & \frac{1}{4} (2 a + 3 c) = \frac{1}{4} \vec{O B} \end{array}$

\vec{O P} = \frac{1}{4} \vec{O B}

therefore OP is parallel to OB and so P must lie on the line OB

If $\begin{array}{rcl} \vec{O P} & = & \frac{1}{4} \vec{O B} \end{array}$ then $\begin{array}{rcl} \vec{P B} & = & \frac{3}{4} \vec{O B} \end{array}$

$\vec{O P} : \vec{P B} = 1 : 3$

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